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Question

# If first three terms of sequence $\frac{1}{16},a,b,\frac{1}{6}$ are in geometric series and last three terms are in harmonic series, then the value of $a$ and $b$ will be

A

$a=-\frac{1}{4},b=1$

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B

$a=\frac{1}{12},b=\frac{1}{9}$

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C

$A$ and $B$ both are true

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D

None of these

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Solution

## The correct option is C $A$ and $B$ both are trueExplanation for the correct option:Step 1. Find the value of $a$ and $b$:Given, $\frac{1}{16},a,b$ are in GP$⇒$${a}^{2}=\frac{b}{16}$ ….(1)Also, $a,b,\frac{1}{6}$ are in HP$⇒$$b=\frac{\frac{2a}{6}}{a+\frac{1}{6}}$$⇒$$b=\frac{2a}{6a+1}$ ….(2)Step 2. Substitute the value of $b$ in equation (1), we get ${a}^{2}=\frac{2a}{16\left(6a+1\right)}$$⇒$ ${a}^{2}=\frac{a}{8\left(6a+1\right)}$$⇒$ $a=\frac{1}{8\left(6a+1\right)}$$⇒$ $48{a}^{2}+8a=1$$⇒$ $48{a}^{2}+8a-1=0$$⇒$$\left(12a-1\right)\left(4a+1\right)=0$$\mathbf{\therefore }\mathbit{a}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{12}}\mathbf{,}\mathbf{}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}$Step 3. Find the value of $b$ by putting the values of $a$:When $a=\frac{1}{12}$, $\begin{array}{rcl}b& =& \frac{2\left(\frac{1}{12}\right)}{6\left(\frac{1}{12}\right)+1}\\ & =& \frac{2}{18}\\ & =& \frac{1}{9}\end{array}$When $a=-\frac{1}{4}$,$\begin{array}{rcl}b& =& \frac{2\left(-\frac{1}{4}\right)}{6\left(-\frac{1}{4}\right)+1}\\ & =& \frac{-2}{-2}\\ & =& 1\end{array}$$\mathbf{\therefore }\mathbit{b}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{9}}\mathbf{,}\mathbf{}\mathbf{1}$Hence, Option ‘C’ is Correct.

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